Problem: The lifespans of lizards in a particular zoo are normally distributed. The average lizard lives $3.1$ years; the standard deviation is $0.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lizard living less than $3.5$ years.
Answer: $3.1$ $2.9$ $3.3$ $2.7$ $3.5$ $2.5$ $3.7$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $3.1$ years. We know the standard deviation is $0.2$ years, so one standard deviation below the mean is $2.9$ years and one standard deviation above the mean is $3.3$ years. Two standard deviations below the mean is $2.7$ years and two standard deviations above the mean is $3.5$ years. Three standard deviations below the mean is $2.5$ years and three standard deviations above the mean is $3.7$ years. We are interested in the probability of a lizard living less than $3.5$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the lizards will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the lizards will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $2.7$ years and the other half $({2.5\%})$ will live longer than $3.5$ years. The probability of a particular lizard living less than $3.5$ years is ${95\%} + {2.5\%}$, or $97.5\%$.